A particle is moving in simple harmonic motion with period 10 seconds and an amplitude of 8 m - HSC - SSCE Mathematics Extension 2 - Question 5 - 2024 - Paper 1

The displacement ( x ) in simple harmonic motion is given by the equation: [ x = A \cos(\omega t) ] where ( A ) is the amplitude and ( t ) is the time.

Substituting the given values:

[ x = 8 \cos\left(\frac{\pi}{5} \times 7.5\right) ] Calculating ( \frac{\pi}{5} \times 7.5 = \frac{15\pi}{10} = \frac{3\pi}{2} ). Hence:

[ x = 8 \cos\left(\frac{3\pi}{2}\right) = 8 \times 0 = 0 \text{ m} ]

The velocity ( v ) in simple harmonic motion is given by: [ v = -A\omega \sin(\omega t) ] Using ( A = 8 ) m and ( \omega = \frac{\pi}{5} ):

[ v = -8 \times \frac{\pi}{5} \sin\left(\frac{3\pi}{2}\right) ] Since ( \sin\left(\frac{3\pi}{2}\right) = -1:\n[ v = -8 \times \frac{\pi}{5} \times -1 = \frac{8\pi}{5} ext{ m/s} ]

After 7.5 seconds, the particle is at the central point of motion with a velocity of ( 0 \text{ m/s} ). Therefore, the correct answer is:

D. 8 m to the right of the central point of motion with a velocity of 0 m s⁻¹.